Thread: Tangent line and Perpendicular

Hello

Determine the coordinates of the point on the graph f(x) = (2x+1)^0.5 where the tangent line is perpendicular to the line 3x +y +4 = 0

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This is what I did so far
f'(x) = 0.5 (2x+1) ^-0.5 (2)
= 1 / (2x+1) ^0.5

Not sure what to do after

Originally Posted by jamson254

Hello

Determine the coordinates of the point on the graph f(x) = (2x+1)^0.5 where the tangent line is perpendicular to the line 3x +y +4 = 0

----------------------------------
This is what I did so far
f'(x) = 0.5 (2x+1) ^-0.5 (2)
= 1 / (2x+1) ^0.5

Not sure what to do after

find the slope of the line, m ... the perpendicular slope is -1/m

set f'(x) = -1/m , then solve for x ... finally, evaluate the y-coordinate from f(x).

Thanks is this correct?
3x+y+4=0
Y = -3x-4

Slope = -3

Perpendicular slope = 1/3

Let F' = 1/3

1/3 = 1/ (2x+1)^0.5



3 = (2x+1) ^.5
squaring both sides
9 = 2x+1
x = 4

F(4) = 3

Point is (4,3)

correct

Thanks a lot for helping me